csc-x-dx-

Question Number 85456 by jagoll last updated on 22/Mar/20

\int \sqrt{\csc x} dx ?

Commented by jagoll last updated on 22/Mar/20

thank sir

Commented by MJS last updated on 22/Mar/20

sadly this is wrong .

t = \sqrt{\sin x} \to d x = \frac{2 \sqrt{\sin x}}{\cos x} d t = \frac{2 t}{\sqrt{1 - t^{4}}} d t

\Rightarrow 2 \int \frac{d t}{\sqrt{1 - t^{4}}} and we {can}^{'} t solve this

Commented by MJS last updated on 22/Mar/20

you should a l w a y s test your solutions

\frac{d}{d x} [2 \arcsin \sqrt{\sin x}] = \frac{\cos x}{\sqrt{\sin x - \sin^{2} x}} \neq \frac{1}{\sqrt{\sin x}}

Commented by jagoll last updated on 22/Mar/20

so what the correct answer sir ?

Commented by MJS last updated on 22/Mar/20

I cannot solve it

{there}^{'} s a solution using Γ (x) somehow, it had

been on this forum, maybe a year ago, I {can}^{'} t

find it

Commented by jagoll last updated on 22/Mar/20

thank you sir

Commented by MJS last updated on 22/Mar/20

found this :

\int \frac{d x}{\sqrt{\sin x}} =

[t = x - \frac{π}{2} \to d x = d t]

\int \frac{d t}{\sqrt{\cos t}} = \int \frac{d t}{\sqrt{1 - {2 \sin}^{2} \frac{t}{2}}} =

= F (\frac{t}{2} ∣ 2) = F (\frac{x}{2} - \frac{π}{4} ∣ 2) + C

{it}^{'} s an eliptic integral

but I cannot explain, {it}^{'} s not my solution

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